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Given matrix A, is it possible to conveniently get the invertible matrices P and Q that appear in the matrix equivalence and satisfy A=P[I_r,0;0,0]Q with numpy or scipy?

Ziyuan
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  • Do you know to calculate the decomposition without python? If yes [link](http://docs.scipy.org/doc/scipy/reference/linalg.html) should give you the needed tools, to implement that in python. – Dschoni Oct 04 '13 at 11:49
  • And what do you mean by [I_r,0;0,0]? – Dschoni Oct 04 '13 at 11:54
  • @Dschoni It's a block matrix, which is almost an identity matrix but with only `r` 1's – Ziyuan Oct 04 '13 at 13:06
  • @Dschoni I can do it as the manual way but I wonder whether there are built-in methods. This it important for calculating all generalized matrix inverses. – Ziyuan Oct 04 '13 at 13:51

1 Answers1

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Probably you can do a singular value decomposition (SVD) and then multiply the singular values to one of the unitary matrice.

See definition of SVD at http://en.wikipedia.org/wiki/Singular_value_decomposition

basically you will get a unitary matrix, a diagonal matrix and another unitary matrix. The diagonal matrix has number of nonzero elements equal to rank(A).

In python code, it's

P, S, Q = numpy.linalg.svd(A)
for i, row in enumerate(Q):
  row *= S[i]

See http://docs.scipy.org/doc/numpy/reference/generated/numpy.linalg.svd.html about numpy.linalg.svd

P, Q then become exactly what you want. But keep in mind there's not a unique way of such decomposition.

Bo-Xiao Zheng
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